HDU 2608 0 or 1（数学推导）

0 or 1

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3273 Accepted Submission(s): 914

Problem Description

Solving problem is a interesting thing. Yifenfei like to slove different problem,because he think it is a way let him more intelligent. But as we know,yifenfei is weak in math. When he come up against a difficult math problem, he always try to get a hand. Now the problem is coming! Let we

define T(n) as the sum of all numbers which are positive integers can divied n. and S(n) = T(1) + T(2) + T(3)…..+T(n).

Input

The first line of the input contains an integer T which means the number of test cases. Then T lines follow, each line consists of only one positive integers n. You may assume the integer will not exceed 2^31.

Output

For each test case, you should output one lines of one integer S(n) %2. So you may see the answer is always 0 or 1 .

Sample Input

3

1

2

3

Sample Output

1

0

0

Hint

Hint S(3) = T(1) + T(2) +T(3) = 1 + (1+2) + (1+3) = 8

S(3) % 2 = 0

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本题是一道比较难的数学推导题目，其实想到了也就那样。

（一）.假如一个数n是一个奇数。

1.假如n不是某个数的平方，那么T（n）一定为0，因为出现了有限对因子。

2.假如n是某个数的平方的话，那么T（n）一定为1，此时i*i=n,那么在有限对因子后多出了一个因子i，所以T(n)为1.

（二）.假如一个数n是一个偶数。

那么这个数一定可以写成2乘以一个奇数的形式，假如2 * i *i=n的话此时T（n）为1，至于分析方式与奇数一样 。

请大家仔细看分析过程，看不懂的话可以打表找下规律。

如有好的见解请留言。

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下面是AC代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

int main()
{
int t,n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
int sum=0;
for(int i=1; i<=n; i++)
{
if(i*i<=n)
sum++;
if(2*i*i<=n)
sum++;
if(i*i>n)
{
break;
}
}
printf("%d\n",sum%2);
}
return 0;
}